### Video Transcript

Find the total pressure at the base of a column with a vertical height of 2.55 meters containing a liquid of density 1,150 kilograms per cubic meter. The top of the column is at sea level. Use a value of 101.3 kilopascals for the atmospheric pressure on the column.

Let’s say that this is our column of liquid, and we’ll label the height of this column ℎ. The very top of this column, we’re told, is at sea level. This means that the pressure of exactly one atmosphere is bearing down on the top of this 2.55-meter-tall column. That pressure is given to us in units of kilopascals. Knowing all this along with the density of the fluid in our column, we want to solve for the pressure at the base of our column of liquid. We’ll call this pressure 𝑃 sub 𝑡 because it’s a total pressure: the pressure due to the column of liquid as well as the atmosphere on top of that column.

If we call the pressure due to the column of liquid 𝑃 sub 𝑐 and that due to the atmosphere 𝑃 sub atm, then we can write an equation that connects 𝑃 sub atm, 𝑃 sub 𝑐, and the total pressure 𝑃 sub 𝑡. That is, 𝑃 sub 𝑡 equals the sum of the other two pressures.

Now, we know the atmospheric pressure — that’s given to us in our problem statement — but we don’t yet know 𝑃 sub 𝑐. However, we can recall the equation that the pressure due to a fluid of density 𝜌 extended a height ℎ equals 𝜌 times 𝑔, the acceleration due to gravity, times ℎ. We’re given the density of the fluid in our column. And let’s call the column’s density 𝜌 sub 𝑐. This means we can write that 𝑃 sub 𝑡, the total pressure, equals 𝜌 sub 𝑐 times 𝑔 times ℎ plus 𝑃 sub atm.

Note that we know the density of the fluid in our column, we know the height ℎ, and we also know that the acceleration due to gravity is 9.8 meters per second squared. All this, along with the fact that 𝑃 sub atm is given as 101.3 kilopascals means that we can now fill in all the values on the right-hand side of this equation.

With those values substituted in, we see that we have one term in our expression for 𝑃 sub 𝑡 here and the other term here. Before we can add these terms, though, we’ll need to make sure that they’re of the same type, that is, that they have the same units.

Looking at the units in the first term, we see that we have meters times meters divided by meters cubed. The factors of meters in the numerator will cancel. And in the denominator, we’ll have simply meters. What we’re finding then is that the overall units of this first term in our expression are kilograms per meter second squared. We now want to make sure that these match up with the units of the second term.

Because the units in our first term are SI units with no prefix, we’ll want the same thing to be true of the units in our second term. That is, we’ll convert the units of this term from kilopascals to pascals. One single kilopascal is 1,000 pascals. So to convert 101.3 kilopascals to pascals, we’ll multiply this number by 1,000. That gives us 101,300 pascals.

Now, for this unit pascal, a pascal is equal to a newton over a meter squared. And a newton is equal to a kilogram meter per second squared. Therefore, we can write a pascal as a kilogram meter per second squared per meter squared. If we divide both the numerator and the denominator of this expression by meter squared, then in the denominator, meter squared cancels out entirely. And in the numerator, one factor of meters divides out. That leaves us with these units.

And we can see now that these units are equal to the units of the first term on the right-hand side of our expression. Since that’s the case, it means we can add these terms together and that we’re ready to calculate 𝑃 sub 𝑡.

This calculation results in 130,038.5 pascals of pressure. For our final answer, we’ll round this result to three significant figures to match the number of significant figures in the height given to us. We can therefore write our answer this way in scientific notation. The total pressure at the bottom of the column of liquid, that is, the pressure due to the column of liquid plus the atmospheric pressure on top, is 1.30 times 10 to the fifth pascals.